1. **Problem statement:** Calculate the probability that the mean BMI of a random sample of 50 healthy 19 to 24-year-olds is greater than or equal to 27.1 kg/m2, given the population mean $\mu = 24.5$ and population standard deviation $\sigma = 9.1$. The sample mean is 27.1 and the sample size is $n=50$. 2. **Formula and explanation:** We use the standard normal distribution to find this probability. The test statistic (z-score) is calculated by: $$z = \frac{\bar{x} - \mu}{SE}$$ where $\bar{x}$ is the sample mean, $\mu$ is the population mean, and $SE$ is the standard error of the mean. 3. **Calculate the standard error (SE):** $$SE = \frac{\sigma}{\sqrt{n}} = \frac{9.1}{\sqrt{50}} \approx 1.29$$ This matches the given standard error. 4. **Calculate the z-score:** $$z = \frac{27.1 - 24.5}{1.29} = \frac{2.6}{1.29} \approx 2.02$$ 5. **Find the probability:** The probability that the sample mean is greater than or equal to 27.1 is the area to the right of $z=2.02$ in the standard normal distribution. Using standard normal tables or a calculator: $$P(Z \geq 2.02) = 1 - P(Z \leq 2.02) \approx 1 - 0.9783 = 0.0217$$ 6. **Interpretation:** The probability is approximately 0.0217, not 0.217. This means there is about a 2.17% chance that the sample mean BMI is 27.1 or higher if the true population mean is 24.5. **Final answer:** $$P(\bar{X} \geq 27.1) \approx 0.0217$$